A multidimensional Hilbert transform approach for barrier option pricing and survival probability calculation

Abstract

This paper proposes a multidimensional Hilbert transform approach for pricing discretely monitored multi-asset barrier options and computing joint survival probability in multivariate exponential Lévy asset price models. We generalize the univariate Hilbert transform method of Feng and Linetsky (Math Financ 18(3), 337–384, 2008) for single-asset barrier options and the well-known Sinc approximation theory of Stenger (Numerical methods based on sinc and analytic functions. Springer, New York, 1993) for computing the one-dimensional Hilbert transform to any dimension. We prove that, for Lévy processes with joint characteristic functions having an exponentially decaying tail, the error of our method decays exponentially in some power of the number of terms used in the expansion for each dimension. Numerical experiments demonstrate the efficiency of our method in the two-dimensional and three-dimensional problems for some popular multivariate Lévy models.

A Proofs

To prove Theorem 2.1, we need the following lemma.

Lemma A.1

Assume (2.5) holds. For any $\pmb {\beta }\in \{0,1\}^n$, $\pmb {\beta }\ne \pmb 0$ and $\pmb {\alpha }\in \{-1,1\}^n$,

$$\begin{aligned}&\left| \frac{\prod _{\{p:\beta _p=1\}}\sin (\pi z_p/h_p)}{(2\pi i)^{|\pmb {\beta }|_1}} \int _{{\mathbb {R}}^{|\pmb {\beta }|_1}} \frac{f(\pmb {\xi }_{\pmb {\beta }} + i\pmb {\alpha }_{\pmb {\beta }} \pmb {d}_{\pmb {\beta }}^- +\pmb {z}_{\pmb {1}-\pmb {\beta }})d\pmb {\xi }_{\pmb {\beta }}}{\prod _{\{j:\beta _j=1\}}(\xi _j+\alpha _jd_j^--z_j)\sin (\pi \zeta _j/h_j)} \right| \\&\quad \le C \prod _{\{j:\beta _j=1\}} \frac{e^{-\pi (d_j-|y_j|)/h_j} (1+ e^{-2\pi |y_j|/h_j}) }{(d_j-|y_j|)(1-e^{-2\pi d_j/h_j}) }. \end{aligned}$$

Proof of Lemma A.1

Applying the following two inequalities

$$\begin{aligned} \sinh (\pi |y|/h)&\le |\sin (\pi (x \pm iy)/h)| \le \cosh (\pi |y|/h), \end{aligned}$$

(A.1)

$$\begin{aligned} |t-z\pm id|&\ge d-|y|\ (z=x+iy,\ |y|<d,\ t\in {\mathbb {R}}), \end{aligned}$$

(A.2)

we obtain that

$$\begin{aligned}&\left| \frac{\prod _{\{p:\beta _p=1\}}\sin (\pi z_p/h_p)}{(2\pi i)^{|\pmb {\beta }|_1}} \int _{{\mathbb {R}}^{|\pmb {\beta }|_1}} \frac{f(\pmb {\xi }_{\pmb {\beta }} + i\pmb {\alpha }_{\pmb {\beta }} \pmb {d}_{\pmb {\beta }}^- +\pmb {z}_{\pmb {1}-\pmb {\beta }})d\pmb {\xi }_{\pmb {\beta }}}{\prod _{\{j:\beta _j=1\}}(\xi _j+\alpha _jd_j^--z_j)\sin (\pi \zeta _j/h_j)} \right| \\&\quad \le \prod _{\{j:\beta _j=1\}} \frac{1}{2\pi (d_j-|y_j|)} \cdot \frac{ \cosh (\pi |y_j|/h_j) }{ \sinh (\pi |d_j|/h_j)} \cdot \int _{{\mathbb {R}}^{|\pmb {\beta }|_1}} \left| f(\pmb {\xi }_{\pmb {\beta }} + i\pmb {\alpha }_{\pmb {\beta }} \pmb {d}_{\pmb {\beta }}^- +\pmb {z}_{\pmb {1}-\pmb {\beta }})\right| d\pmb {\xi }_{\pmb {\beta }}. \end{aligned}$$

Simplifying and using (2.5) give us the desired bound. $\square$

Theorem 2.1

(1) We prove the error bounds by induction. The case with $n=1$ is proved in Stenger (1993). Suppose our error bounds hold for all dimensions that is less than n. For $j=1,\cdots ,n$, let $0< \delta _j < d_j$ and define

$$\begin{aligned} {\mathcal {D}}_j(m_j,\delta _j) = \left\{ z_j \in {\mathbb {C}}: |\text {Re}z_j|<\left( m_j+\frac{1}{2}\right) h_j, \, |\text {Im}z_j|<\delta _j \right\} . \end{aligned}$$

Fix $z_j = x_j + i y_j$ in ${\mathcal {D}}_{d_j}$, and set $\zeta _j = \xi _j + i \eta _j$ ($j=1,\cdots ,n$). If $m_j$ is sufficiently large and $\delta _j$ is sufficiently close to $d_j$, then $z_j \in {\mathcal {D}}_j(m_j,\delta _j)$. And let

$$\begin{aligned} E(\pmb {m}, \pmb \delta , f)(\pmb {z}) = f(\pmb {z}) - \sum _{k_1=-m_1}^{m_1}\cdots \sum _{k_n=-m_n}^{m_n} f(\pmb {kh}) \prod _{j=1}^{n}S(k_j,h_j)(z_j). \end{aligned}$$

(A.3)

We first show by induction that

$$\begin{aligned}&\frac{\prod _{p=1}^n\sin (\pi z_p/h_p)}{(2\pi i)^n} \int _{\partial {\mathcal {D}}_1(m_1,\delta _1)}\cdots \int _{\partial {\mathcal {D}}_n(m_n,\delta _n)}\frac{ f(\pmb \zeta )d\pmb \zeta }{\prod _{j=1}^n(\zeta _j-z_j)\sin (\pi \zeta _j/h_j)} \nonumber \\&\quad = \sum _{\pmb {\beta } \in \{0,1\}^n }(-1)^{|\pmb {\beta }|_1} \underset{-M_p\le k_p\le M_p}{\sum \textstyle ^{\pmb {\beta }} \textstyle } \left[ f(\pmb {k}_{\pmb {\beta }}\pmb {h}_{\pmb {\beta }} + \pmb {z}_{1-\pmb {\beta }}) \cdot \prod _{\{j: \beta _j=1 \}}S(k_j,h_j)(z_j) \right] . \end{aligned}$$

(A.4)

For $n=1$, the result is given by Cauchy’s residue theorem. Now suppose the equality holds for the $n-1$ case, then

$$\begin{aligned}&\frac{\prod _{p=1}^n\sin (\pi z_p/h_p)}{(2\pi i)^n} \int _{\partial {\mathcal {D}}_1(m_1,\delta _1)}\cdots \int _{\partial {\mathcal {D}}_n(m_n,\delta _n)}\frac{ f(\pmb \zeta )d\pmb \zeta }{\prod _{j=1}^n(\zeta _j-z_j)\sin (\pi \zeta _j/h_j)} \\&\quad = \frac{\prod _{p=1}^{n-1}\sin (\pi z_p/h_p)}{(2\pi i)^{n-1}} \int _{\partial {\mathcal {D}}_1(m_1,\delta _1)}\cdots \int _{\partial {\mathcal {D}}_{n-1}(m_{n-1},\delta _{n-1})} \frac{d\zeta _{1} \cdots d \zeta _{n-1} }{\prod _{j=1}^{n-1}(\zeta _j-z_j)\sin (\pi \zeta _j/h_j)} \\&\qquad \cdot \frac{\sin (\pi z_n/h_n)}{2\pi i}\int _{\partial {\mathcal {D}}_n(m_n,\delta _n)}\frac{f( \zeta _{1},\cdots ,\zeta _{n-1},\zeta _n) d\zeta _n}{(\zeta _n-z_n)\sin (\pi \zeta _n/h_n)} \\&\quad = \frac{\prod _{j=1}^{n-1}\sin (\pi z_j/h_j)}{(2\pi i)^{n-1}} \int _{\partial {\mathcal {D}}_1(m_1,\delta _1)}\cdots \int _{\partial {\mathcal {D}}_{n-1}(m_{n-1},\delta _{n-1})} \frac{d\zeta _{1} \cdots d \zeta _{n-1} }{\prod _{j=1}^{n-1}(\zeta _j-z_j)\sin (\pi \zeta _j/h_j)} \\&\qquad \cdot \left[ f(\zeta _{1},\cdots ,\zeta _{n-1}, z_n)- \sum _{k_n=-m_n}^{m_n}\frac{f(\zeta _{1},\cdots ,\zeta _{n-1},k_nh_n)(-1)^{k_n}}{\pi (z_n-k_nh_n)/h_n}\sin (\pi z_n/h_n)\right] \\&\quad = \sum _{\pmb {\beta }' \in \{0,1\}^{n-1} }(-1)^{|\pmb {\beta }'|_1} \underset{-M_p\le k_p'\le M_p}{\sum \textstyle ^{\pmb {\beta }' \textstyle }} \left[ f(\pmb {k}'_{\pmb {\beta }'} \pmb {h}'_{\pmb {\beta }'} + \pmb {z}'_{\pmb {1}-\pmb {\beta }'},z_n ) \cdot \prod _{\{j: \beta _j'=1 \}}S(k_j,h_j)(z_j) \right] \\&\qquad -\sum _{k_n=-m_n}^{m_n} S(k_n,h_n)(z_n) \sum _{\pmb {\beta }' \in \{0,1\}^{n-1} }(-1)^{|\pmb {\beta }'|_1} \underset{-M_p\le k_p'\le M_p}{\sum \textstyle ^{\pmb {\beta }' \textstyle }} \\&\qquad \quad \left[ f(\pmb {k}'_{\pmb {\beta }'} \pmb {h}'_{\pmb {\beta }'} + \pmb {z}'_{\pmb {1}-\pmb {\beta }'},k_nh_n )\prod _{\{j: \beta _j'=1 \}}S(k_j,h_j)(z_j) \right] \\&\quad = \sum _{\pmb {\beta } \in \{0,1\}^n }(-1)^{|\pmb {\beta }|_1} \left[ \underset{-M_p\le k_p\le M_p}{\sum \textstyle ^{\pmb {\beta } \textstyle }} \left( f(\pmb {k}_{\pmb {\beta }} \pmb {h}_{\pmb {\beta }} + \pmb {z}_{\pmb {1}-\pmb {\beta }} )\prod _{\{j: \beta _j=1 \}}S(k_j,h_j)(z_j) \right) \right] . \end{aligned}$$

The third equality follows from induction, the fourth equality is obtained from rewriting $\pmb {z} = (\pmb {z}',z_n),\pmb {k} = (\pmb {k}',k_n),\pmb {h} = (\pmb {h}',h_n)$. Thus by induction, we obtain (A.4).

Noting that the number of positive and negative terms in $\{(-1)^{|\pmb {\beta }|_1}:\pmb {\beta }\in \{0,1\}^n\}$ are the same, we can rewrite the right side of (A.4) as

$$\begin{aligned} \sum _{\pmb {\beta } \in \{0,1\}^n }(-1)^{|\pmb {\beta }|_1+1} \left[ f(\pmb {z}) - \underset{-M_p\le k_p\le M_p}{\sum \textstyle ^{\pmb {\beta } \textstyle }} \left( f(\pmb {k}_{\pmb {\beta }} \pmb {h}_{\pmb {\beta }} + \pmb {z}_{\pmb {1}-\pmb {\beta }} )\prod _{\{j: \beta _j=1 \}}S(k_j,h_j)(z_j) \right) \right] . \end{aligned}$$

(A.5)

Thus using (A.5), the term $E(\pmb {m}, \pmb \delta , f)(\pmb {z})$ that is defined in (A.3) can be written as

$$\begin{aligned}&E(\pmb {m}, \pmb \delta , f)(\pmb {z}) =(-1)^{n+1} \frac{\prod _{p=1}^n\sin (\pi z_p/h_p)}{(2\pi i)^n} \nonumber \\&\quad \int _{\partial {\mathcal {D}}_1(m_1,\delta _1)}\cdots \int _{\partial {\mathcal {D}}_n(m_n,\delta _n)} \frac{ f(\pmb \zeta )d\pmb \zeta }{\prod _{j=1}^n(\zeta _j-z_j)\sin (\pi \zeta _j/h_j)} \nonumber \\&\quad + \sum _{\pmb {\beta } \in \{0,1\}^n, \pmb {\beta } \ne \pmb {1} }(-1)^{|\pmb {\beta }|_1+n+1} \left[ f(\pmb {z}) - \underset{-M_p\le k_p\le M_p}{\sum \textstyle ^{\pmb {\beta } \textstyle }} \nonumber \right. \\&\left. \qquad \left( f(\pmb {k}_{\pmb {\beta }} \pmb {h}_{\pmb {\beta }} + \pmb {z}_{\pmb {1}-\pmb {\beta }})\cdot \prod _{\{j: \beta _j=1 \}}S(k_j,h_j)(z_j) \right) \right] . \end{aligned}$$

(A.6)

We now derive the limit of the integral in (A.6) by taking $m_j$ to $\infty$ for all j. The integral can be split into contributions from the product of horizontal segments only, those from the product of vertical segments only and those from the mixed product of horizontal and vertical segments. We observe that whenever there is a vertical segment involved in the integration region, the integral vanishes in the limit. We illustrate this result by considering an integral of this type below, where a vertical segment is used for the n-th dimension and horizontal segments are used in all other dimensions. Here $\zeta _j = \xi _j+id_j, 1\le j\le n-1, \zeta _n:= \left( m_n+\frac{1}{2} \right) h_n+i\eta _n$, and let $\tilde{\pmb {\xi }} = (\xi _1,\cdots ,\xi _{n-1})' , \tilde{\pmb {d}} = (d_1,\cdots , d_{n-1} )'$.

$$\begin{aligned}&\left| \int _{-\delta _{n}}^{\delta _{n}} \int _{-\left( m_1+\frac{1}{2}\right) h_1}^{\left( m_1+\frac{1}{2}\right) h_1} \cdots \int _{-\left( m_{n-1}+\frac{1}{2}\right) h_{n-1}}^{\left( m_{n-1}+\frac{1}{2}\right) h_{n-1} } \frac{ f\left( \tilde{\pmb {\xi }} + i \tilde{\pmb {d}}^- , \left( m_n+\frac{1}{2} \right) h_n+i\eta _n \right) d\tilde{\pmb {\xi }}d\eta _n }{\prod _{j=1}^{n-1}(\zeta _j-z_j)\sin (\pi \zeta _j/h_j) \cdot (\zeta _n-z_n) \sin (\pi \zeta _n/h_n)} \right| \\&\le \int _{-\delta _{n}}^{\delta _{n}} \left| \int _{-\left( m_1+\frac{1}{2}\right) h_1}^{\left( m_1+\frac{1}{2}\right) h_1} \cdots \int _{-\left( m_{n-1}+\frac{1}{2}\right) h_{n-1}}^{\left( m_{n-1}+\frac{1}{2}\right) h_{n-1} } \frac{ f(\tilde{\pmb {\xi }} + i \tilde{\pmb {d}}^-, \left( m_n+\frac{1}{2} \right) h_n+i\eta _n ) d\tilde{\pmb {\xi }} }{\prod _{j=1}^{n-1}(\xi _j+id_j-z_j)\sin (\pi (\xi _j+id_j)/h_j) } \right| d\eta _n \\&\qquad \cdot \frac{1}{\left| \left( m_n+\frac{1}{2}\right) h_n-x_n \right| } \\&\rightarrow 0 \text { as } m_j \rightarrow \infty \ \text {for}\ j=1,\cdots ,n. \end{aligned}$$

The inequaltiy is obtained as follows. Along $\{(m_n+\frac{1}{2})h_n+i\eta _n: -\delta _n\le \eta _n\le \delta _n\}$, we have

$$\begin{aligned}&|\zeta _n-z_n| \ge \left| \left( m_n+\frac{1}{2}\right) h_n-x_n \right| , \\&|\sin (\pi \zeta _n/h_n) | = | (-1)^n \cosh (\pi \eta _n/h_n)| \ge 1. \end{aligned}$$

In addition, the $(n-1)$-dimensional integral can be bounded using Lemma A.1.

Putting these limiting results together, when $m_j\rightarrow \infty$, $\delta _j\rightarrow d_j$ for all j, the first part of the RHS of (A.6) becomes

$$\begin{aligned}&\frac{\prod _{p=1}^n\sin (\pi z_p/h_p)}{(2\pi i)^n} \int _{\partial {\mathcal {D}}_1(m_1,\delta _1)}\cdots \int _{\partial {\mathcal {D}}_n(m_n,\delta _n)}\frac{ f(\pmb \zeta )d\pmb \zeta }{\prod _{j=1}^n(\zeta _j-z_j)\sin (\pi \zeta _j/h_j)} \nonumber \\&\quad =\frac{\prod _{p=1}^n\sin (\pi z_p/h_p)}{(2\pi i)^n}\nonumber \\&\quad \sum _{\pmb {\alpha } \in \{-1,1\}^n } (-1)^{\sum _{\{j:\alpha _j=1\}}1}\int _{{\mathbb {R}}^n} \frac{f(\pmb {\xi } + i\pmb {\alpha } \pmb {d}^- )d\pmb {\xi }}{\prod _{j=1}^n(\xi _j+i\alpha _jd_j^--z_j)\sin (\pi \zeta _j/h_j)} . \end{aligned}$$

(A.7)

The RHS of (A.7) can be bounded as

$$\begin{aligned} C\prod _{ j=1}^n \frac{e^{-\pi (d_j-|y_j|)/h_j} (1+ e^{-2\pi |y_j|/h_j}) }{(d_j-|y_j|)(1-e^{-2\pi d_j/h_j})} \end{aligned}$$

(A.8)

using Lemma A.1. For the second part of the RHS of (A.6), applying the induction assumption together with (2.5) and (2.6), we obtain for each $\pmb {\beta } \in \{0,1\}^n, \pmb {\beta } \ne \pmb 0, \pmb {1}$,

$$\begin{aligned}&\left| f(\pmb {z}) - \underset{-M_j\le k_j\le M_j}{\sum \textstyle ^{\pmb {\beta } \textstyle }} \left( f(\pmb {k}_{\pmb {\beta }} \pmb {h}_{\pmb {\beta }} + \pmb {z}_{\pmb {1}-\pmb {\beta }})\cdot \prod _{\{j: \beta _j=1 \}}S(k_j,h_j)(z_j) \right) \right| \nonumber \\&\quad \le \sum _{\pmb {\alpha } \in \{0,1\}^n, \pmb {\alpha } \le \pmb {\beta }, \pmb {\alpha } \ne 0} \left[ C_{\pmb {\alpha }} \prod _{\{p:\alpha _p =1 \}}\frac{e^{-\pi (d_p-|y_p|)/h_p}(1+e^{-2\pi |y_p|/h_p})}{(d_p-|y_p|)(1-e^{-2\pi d_p/h_p})} \right] . \end{aligned}$$

(A.9)

Here $\pmb {\alpha } \le \pmb {\beta }$ means $\alpha _j \le \beta _j$ for all $1\le j \le n$. Adding (A.8) and (A.9) over $\pmb {\beta }\ne \pmb 0,\pmb 1$ yields (2.7).

(2) We have $E_H(f,\pmb {h})(\pmb {\xi })=\frac{1}{\pi ^n} \int _{{\mathbb {R}}^n}\frac{E(f,\pmb {h})(\pmb {x})}{\prod _{i=1}^{n}(\xi _i - x_i)}d\pmb {x}$. Thus (2.8) can be derived by using (A.7) with $z_i=x_i$, interchanging the order of integration and applying the identity

$$\begin{aligned} \frac{1}{\pi }\int _{{\mathbb {R}}}\frac{\sin (\pi x/h)}{(\omega -x)(z-x)}dx=\frac{e^{i\pi \omega \text {sgn}(\text {Im}(\omega ))/h}-e^{i\pi z\text {sgn}(\text {Im}(z))/h}}{\omega -z} \end{aligned}$$

together with (A.1) and (A.2).

(3) We have $E_I(f,\pmb {h})=\int _{{\mathbb {R}}^n}E(f,\pmb {h})(\pmb {x})d\pmb {x}$. Using (A.7) with $z_i=x_i$, interchanging the order of integration and applying the identities

$$\begin{aligned} \frac{1}{2\pi i}\int _{{\mathbb {R}}}\frac{\sin (\pi x/h)}{t-x-id}dx=\frac{i}{2}e^{-\pi (d+it)/h},\ \frac{1}{2\pi i}\int _{{\mathbb {R}}}\frac{\sin (\pi x/h)}{x-t-id}dx=\frac{i}{2}e^{-\pi (d-it)/h}, \end{aligned}$$

and (A.1), we obtain (2.9).

$\square$

Proposition 3.2

It holds that

$$\begin{aligned} E_{\pmb {x}}[ f(\pmb {X}_t)] = E_{\pmb {x}}[ e^{-\pmb {\alpha }' \pmb {X}_t}f_{\pmb {\alpha }}(\pmb {X}_t) ]=e^{-\pmb {\alpha }'\pmb {x}-t\psi (i\pmb {\alpha })}E_{\pmb {x}}^{(\pmb {\alpha })}[ f_{\pmb {\alpha }}(\pmb {X}_t)]. \end{aligned}$$

Take the Fourier transform of $E_{\pmb {x}}^{(\pmb {\alpha })}\left[ f_{\pmb {\alpha }}(\pmb {X}_t) \right]$ as a function of $\pmb {x}$ and use Proposition 9 in Bertoin (1998), we obtain

$$\begin{aligned} {\mathcal {F}}_n\left( E_{\pmb {x}}^{(\pmb {\alpha })}\left[ f_{\pmb {\alpha }}(\pmb {X}_t) \right] \right) (\pmb {\xi })=\phi _t^{\pmb {\alpha }}(-\pmb {\xi }){\hat{f}}_{\pmb {\alpha }}(\pmb {\xi }). \end{aligned}$$

Due to the integrability condition (3.6), we can invert the Fourier transform and obtain (3.8). $\square$

Proposition 3.3

Given $\pmb {\beta } \in \{0,1\}^n$, let $k_{\pmb {\beta }}(\pmb {\xi }) = \prod _{\{j:\beta _j=1 \}}1/(\pi \xi _j)$. The partial Hilbert transform ${\mathcal {H}}_{\pmb {\beta }}$ of ${\hat{f}}$ is the convolution of ${\hat{f}}$ and $k_{\pmb {\beta }}$, i.e. ${\mathcal {H}}_{\pmb {\beta }}{\hat{f}}(\pmb {\xi }) = ({\hat{f}} *k_{\pmb {\beta }})(\pmb {\xi })$ (see Eq.(15.39) in King 2009). The convolution theorem shows that

$$\begin{aligned}&{\mathcal {F}}^{-1}_{\pmb {\beta }}\left( {\mathcal {H}}_{\pmb {\beta }}{\hat{f}}(\pmb {\xi })\right) (\pmb {x}_{\pmb {\beta }}+\pmb {\xi }_{\pmb {1}-\pmb {\beta }})\nonumber \\&\quad = {\mathcal {F}}^{-1}_{\pmb {\beta }}({\hat{f}}(\pmb {\xi }))(\pmb {x}_{\pmb {\beta }}+\pmb {\xi }_{\pmb {1}-\pmb {\beta }}) \cdot (2\pi )^{|\pmb {\beta }|_1}{\mathcal {F}}^{-1}_{\pmb {\beta }}(k_{\pmb {\beta }}(\pmb {\xi }))(\pmb {x}_{\pmb {\beta }}+\pmb {\xi }_{\pmb {1}-\pmb {\beta }}), \end{aligned}$$

(A.10)

where

$$\begin{aligned}&{\mathcal {F}}^{-1}_{\pmb {\beta }}(k_{\pmb {\beta }}(\pmb {\xi }))(\pmb {x}_{\pmb {\beta }}+\pmb {\xi }_{\pmb {1}-\pmb {\beta }}) = \left( \frac{1}{2\pi } \right) ^{|\pmb {\beta }|_1}\\&\quad \int _{{\mathbb {R}}^{|\pmb {\beta }|_1}}\prod _{\{j:\beta _j=1 \}}\frac{e^{-i x_j\xi _j}d\xi _j}{\pi \xi _j}= \left( \frac{-i}{2\pi } \right) ^{|\pmb {\beta }|_1} \prod _{\{j:\beta _j=1 \}}\text {sgn}(x_j). \end{aligned}$$

The function $k_{\pmb {\beta }}(\pmb {\xi })$ is not in $L^1({\mathbb {R}}^n)$, so ${\mathcal {F}}^{-1}_{\pmb {\beta }}(k_{\pmb {\beta }}(\pmb {\xi }))(\pmb {x}_{\pmb {\beta }}+\pmb {\xi }_{\pmb {1}-\pmb {\beta }})$ must be interpreted as a Cauchy principal integral. We further apply Fourier inversion to those dimensions p with $\beta _p=0$ on both sides of (A.10) and obtain that

$$\begin{aligned} {\mathcal {F}}^{-1}_n\left( {\mathcal {H}}_{\pmb {\beta }}{\hat{f}}(\pmb {\xi })\right) (\pmb {x}) = i^{|\pmb {\beta }|_1} f(\pmb {x})\prod _{\{j:\beta _j=1 \}}\text {sgn}(x_j). \end{aligned}$$

Here, we use the identity that ${\mathcal {F}}^{-1}_{\pmb {1}-\pmb {\beta }}({\mathcal {F}}^{-1}_{\pmb {\beta }}({\hat{f}}(\pmb {\xi }))(\pmb {x}_{\pmb {\beta }}+\pmb {\xi }_{\pmb {1}-\pmb {\beta }}))(\pmb {x})={\mathcal {F}}^{-1}_n{\hat{f}}(\pmb {x})=f(\pmb {x})$ as ${\hat{f}}\in L^1({\mathbb {R}}^n,{\mathbb {C}})$. Further applying ${\mathcal {F}}_n$ on both sides of the above equation gives us (3.10). $\square$

Proposition 3.4

Consider the translation operators

$$\begin{aligned} \left( {\mathcal {T}}_{l_j}^{(j)}f\right) (\pmb {x})&:= f(x_1,\cdots ,x_{j-1},x_j-l_j,x_{j+1},\cdots , x_n). \end{aligned}$$

We can write

$$\begin{aligned} \mathbbm {1}_{(\pmb {l},\infty )}(\pmb {x})f(\pmb {x})= & {} \frac{1}{2^n} \prod _{j=1}^{n} \left( 1+ {\mathcal {T}}_{l_j}^{(j)}\text {sgn}(x_j) \right) f(\pmb {x})\\= & {} \frac{1}{2^n} \sum _{\pmb {\beta } \in \{0,1\}^n } f(\pmb {x}) \cdot {\mathcal {T}}_{\pmb {l}}^{\pmb {\beta }}\left( \prod _{ \{ p:\beta _p = 1\}} \text {sgn}(x_p) \right) , \end{aligned}$$

where ${\mathcal {T}}_{\pmb {l}}^{\pmb {\beta }} = \prod _{ \{ p:\beta _p = 1\}} {\mathcal {T}}_{l_p}^{(p)}$. Furthermore, we have

$$\begin{aligned}f(\pmb {x}) \cdot {\mathcal {T}}_{\pmb {l}}^{\pmb {\beta }}\left( \prod _{ \{ p:\beta _p = 1\}} \text {sgn}(x_p) \right) = {\mathcal {T}}_{\pmb {l}}^{\pmb {\beta }} \left( \prod _{ \{ p:\beta _p = 1\}} \text {sgn}(x_p) \cdot {\mathcal {T}}_{-\pmb {l}}^{\pmb {\beta }}f(\pmb {x}) \right) . \end{aligned}$$

Then, by (3.10) and the property of the Fourier transform w.r.t. translation, we obtain

$$\begin{aligned} {\mathcal {F}}_n\left( {\mathcal {T}}_{\pmb {l}}^{\pmb {\beta }} \left( \prod _{ \{ p:\beta _p = 1\}} \text {sgn}(x_p) \cdot {\mathcal {T}}_{-\pmb {l}}^{\pmb {\beta }}f(\pmb {x}) \right) \right) (\pmb {\xi }) = i^{|\pmb {\beta }|} e^{i \pmb {\xi }_{\pmb {\beta }}' \pmb {l}_{\pmb {\beta } }} {\mathcal {H}}_{\pmb {\beta }} \left( e^{-i \pmb {\eta }_{\pmb {\beta }}' \pmb {l}_{\pmb {\beta } }}{\hat{f}}\left( \pmb {\eta }_{\pmb {\beta }}+\pmb {\xi }_{\pmb {1}-\pmb {\beta }}\right) \right) (\pmb {\xi }). \end{aligned}$$

Applying the ${\mathcal {F}}_n$ to $\mathbbm {1}_{(\pmb {l},\infty )}(\pmb {x})f(\pmb {x})$ and using the above results gives us (3.11). $\square$

Theorem 3.1

First, notice that if (3.18) holds, then for any $t\ge \chi$,

$$\begin{aligned} \int _{{\mathbb {R}}^n}\big |\phi _{t}^{\pmb {\alpha }}(\pmb {\xi }) \big |d\pmb {\xi }=\int _{{\mathbb {R}}^n}\big |\phi _{\chi }^{\pmb {\alpha }}(\pmb {\xi })\phi _{t-\chi }^{\pmb {\alpha }}(\pmb {\xi }) \big |d\pmb {\xi }\le \int _{{\mathbb {R}}^n}\big |\phi _{\chi }^{\pmb {\alpha }}(\pmb {\xi })\big |d\pmb {\xi }<\infty . \end{aligned}$$

We prove that (3.19) holds for every j. For $j=N$, it is given by the assumption (3.17). Now assume that the claim (3.19) holds for j to N and we prove below that it also holds for $j-1$. Using the recursion (3.15), we have

$$\begin{aligned} \int _{{\mathbb {R}}^n}\big |\phi _{\Delta }^{\pmb {\alpha }}(-\pmb {\xi }) {\hat{v}}^{j-1}_{\pmb {\alpha }}(\pmb {\xi })\big |d\pmb {\xi } \le \frac{e^{-\Delta \psi (i\pmb {\alpha })}}{2^n}\sum _{\pmb {\beta } \in \{0,1\}^n } I_{\pmb {\beta }}, \end{aligned}$$

where

$$\begin{aligned} I_{\pmb {\beta }} := \int _{{\mathbb {R}}^n} \left| \phi _{\Delta }^{\pmb {\alpha }}(-\pmb {\xi }) {\mathcal {H}}_{\pmb {\beta }} \left( e^{-i \pmb {\eta }_{\pmb {\beta }}' \pmb {l}_{\pmb {\beta } }} \phi _\Delta ^{\pmb {\alpha }} \big (-\pmb {\eta }_{\pmb {\beta }} -\pmb {\xi }_{\pmb {1}-\pmb {\beta }} \big ){\hat{v}}_{\pmb {\alpha }}^j\big (\pmb {\eta }_{\pmb {\beta }}+\pmb {\xi }_{\pmb {1}-\pmb {\beta }}\big ) \right) (\pmb {\xi }) \right| d\pmb {\xi }. \end{aligned}$$

As $|\phi _{\Delta }^{\pmb {\alpha }}(-\pmb {\xi })|$ is bounded over $\pmb {\xi }$, $I_{\pmb {0}}$ is finite by the induction assumption. Now pick $p > 1$ such that $p\Delta \ge \chi$ and set $q=p/(p-1)$. Note that $(\phi _{\Delta }^{\pmb {\alpha }}(-\pmb {\xi }))^p = \phi _{p\Delta }^{\pmb {\alpha }}(-\pmb {\xi }) \in L^1({\mathbb {R}}^n)$, and this implies $\phi _{\Delta }^{\pmb {\alpha }}(-\pmb {\xi }) \in L^p({\mathbb {R}}^n)$. The Calderón–Zygmund inequality (see Eq. (15.115) in King 2009) says that, if $f \in L^p({\mathbb {R}}^n)$ with $p>1$, then

$$\begin{aligned} \Vert {\mathcal {H}}_nf \Vert _p&= \left\| \prod _{k=1}^n {\mathcal {H}}_{(k)}f \right\| _p = \left\| {\mathcal {H}}_{(1)}\prod _{k=2}^n {\mathcal {H}}_{(k)}f \right\| _p\\&\le C_p^{(1)} \left\| \prod _{k=2}^n {\mathcal {H}}_{(k)}f \right\| _p \le \cdots \le C_p^{(n)} \left\| f\right\| _p, \end{aligned}$$

for some constants $C_p^{(k)},1\le k\le n$. Here, we write ${\mathcal {H}}_{(k)}$ as the partial Hilbert transform with $\pmb {\beta } = (0,\cdots ,0,1,0,\cdots ,0)$, where only $\beta _k=1$. Thus for $I_{\pmb {1}}$, by Hölder’s inequality,

$$\begin{aligned} I_{\pmb {1}}&\le \Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot )\Vert _p \cdot \left\| {\mathcal {H}}_{n}\left( e^{-i\pmb {\eta }'\pmb {l}}\phi _{\Delta }^{\pmb {\alpha }}(-\pmb {\eta }){\hat{v}}^j_{\pmb {\alpha }}(\pmb {\eta })\right) (\cdot )\right\| _q \le C \Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot )\Vert _p \cdot \Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ){\hat{v}}^j_{\pmb {\alpha }}(\cdot ) \Vert _q. \end{aligned}$$

Using the boundedness of ${\hat{v}}^j_{\pmb {\alpha }}$, we have

$$\begin{aligned} \Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ){\hat{v}}^j_{\pmb {\alpha }}(\cdot ) \Vert _q\le \Vert {\hat{v}}^j_{\pmb {\alpha }}(\cdot )\Vert _{\infty }^{q-1}\Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ){\hat{v}}^j_{\pmb {\alpha }}(\cdot ) \Vert _{1}, \end{aligned}$$

which is finite by the induction assumption. This shows $I_{\pmb {1}}$ is finite. The finiteness of $\{I_{\pmb {\beta }}: \pmb {\beta } \ne \pmb {0},\pmb {1} \}$ can be proved similarly. Finally, (3.20) follows from Proposition 3.2. $\square$

Theorem 3.2

(1) For ${\mathcal {P}}^\Delta g(\pmb {\xi })$, the total approximation error consists of the error for the terms involving several partial Hilbert transforms and the term with the n-dimensional Hilbert transform. We can directly apply Theorem 2.1 to estimate the approximation error for the partial Hilbert transforms under (3.24) and the boundedness of g. Below we only analyze the error for the n-dimensional Hilbert transform. We have

$$\begin{aligned} \left| {\mathcal {P}}^{\Delta }g(\pmb {\xi })-{\mathcal {P}}^{\Delta }_{h,M}g(\pmb {\xi })\right|&\le \left| {\mathcal {P}}^{\Delta }g(\pmb {\xi })-{\mathcal {P}}^{\Delta }_{h,\infty }g(\pmb {\xi })\right| +\left| {\mathcal {P}}^{\Delta }_{h,\infty }g(\pmb {\xi })-{\mathcal {P}}^{\Delta }_{h,M}g(\pmb {\xi })\right| . \end{aligned}$$

The first term can be estimated by applying (2.8) as our assumption on characteristic function (3.24) and the boundedness of g allows us to verify the conditions in Theorem 2.1. For the second term, it is bounded by

$$\begin{aligned}&\Vert g\Vert _{L^\infty ({\mathbb {R}}^n)}\sum _{|m_j|>M_j, 1\le j\le n}\left| \phi _{\Delta }(- \pmb {mh})\right| \nonumber \\&\quad \le a\cdot 2^n\Vert g\Vert _{L^\infty ({\mathbb {R}}^n)} \left( \prod _{j=1}^nh_j^{-1} \right) \sum _{m_j>M_j,1\le j\le n} \left( e^{-\Delta b \left( \sum _{j=1}^{n}m_j^2h_j^2 \right) ^{\nu /2} } \cdot \prod _{j=1}^nh_j\right) \nonumber \\&\quad \le a \cdot 2^n\Vert g\Vert _{L^\infty ({\mathbb {R}}^n)} \left( \prod _{j=1}^nh_j^{-1} \right) \sum _{m_j>M_j,1\le j\le n} e^{-\Delta b n^{\nu /2-1}\sum _{j=1}^{n}\left( m_jh_j \right) ^{\nu }}\cdot \prod _{j=1}^nh_j \nonumber \\&\quad \le a\cdot 2^n\Vert g\Vert _{L^\infty ({\mathbb {R}}^n)} \left( \prod _{j=1}^nh_j^{-1} \right) \cdot \prod _{j=1}^n \int _{M_jh_j}^\infty e^{-\Delta bn^{\nu /2-1}x_j^{\nu }}dx_j\nonumber \\&\quad \le a\cdot 2^n\Vert g\Vert _{L^\infty ({\mathbb {R}}^n)} \prod _{j=1}^{n} \Gamma \left( 1/\nu ,\Delta bn^{\nu /2-1}(M_jh_j)^\nu \right) /h_j, \end{aligned}$$

(A.11)

where we used $|(1-\cos (x))/x|{\le } 1$, (3.23), $\left( \sum _{j{=}1}^{n}m_j^2h_j^2\right) ^{\nu /2}\ge n^{\nu /2-1}\sum _{j{=}1}^{n}(m_jh_j)^\nu$ (due to the concavity of $x^{\nu /2}$ as $\nu \in (0,2]$). Combining Eq.(6.20) in Feng and Linetsky (2008a), (2.8) and (A.11), we arrive at (3.26). Now we set $h_j$ according to (3.27) and $h_j$ is bounded as $M_j\ge 1$. Therefore, for some constant $C>0$,

$$\begin{aligned} \frac{e^{-\pi d_j/h_j}}{1-e^{-\pi d_j/h_j}}\le Ce^{-\pi d_j/h_j}, \end{aligned}$$

(A.12)

where the constant C is independent of j. As $\Gamma (a,x)\sim x^{a-1}e^{-x}$ for x large, we can bound $\Gamma (a,x)$ as a constant times $x^{a-1}e^{-x}$ for all x bounded away from 0. Using this estimate, (3.27), (A.12) and that the term $M_j^{\frac{2}{1+\nu }}\exp \left( -2cM_j^{\frac{\nu }{1+\nu }}\right) \le A M_j^{\frac{1}{1+\nu }}\exp \left( -cM_j^{\frac{\nu }{1+\nu }}\right)$ for some constant $A>0$ when $c>0$, we obtain (3.28).

(2) The proof for ${\mathcal {R}}^\Delta g(\pmb {x})$ is similar to the proof for ${\mathcal {P}}^\Delta g(\pmb {\xi })$, so the detail is omitted. $\square$

Theorem 3.3

(1) From the proof of Theorem 3.1, we have that for every $q>1$,

$$\begin{aligned} \Vert {\mathcal {H}}_{\pmb {\beta }}f\Vert _{L^q}\le C_q^{\pmb {\beta }}\Vert f\Vert _{L^q}, \end{aligned}$$

(5.13)

for some constant $C_q^{\pmb {\beta }}>0$. Now we use this result to bound $\Vert {\mathcal {P}}^{\Delta } g \Vert _{L^q}$ for $g\in L^q$. Using the expression of ${\mathcal {P}}^{\Delta } g$ yields

$$\begin{aligned} \left\| {\mathcal {P}}^{\Delta }g(\pmb {\xi }) \right\| _{L^q}&= \left\| \sum _{\pmb {\beta } \in \{0,1\}^n } i^{|\pmb {\beta }|_1} e^{i\pmb {\xi }_{\pmb {\beta }}' \pmb {l}_{\pmb {\beta }} }{\mathcal {H}}_{\pmb {\beta }} \left( e^{-i\pmb {\eta }_{\pmb {\beta }}' \pmb {l}_{\pmb {\beta }}}\phi _\Delta ^{\pmb {\alpha }} \big (-\pmb {\eta }_{\pmb {\beta }}-\pmb {\xi }_{\pmb {1}-\pmb {\beta }}\big )g\big (\pmb {\eta }_{\pmb {\beta }}+\pmb {\xi }_{\pmb {1}-\pmb {\beta }} \big ) \right) (\pmb {\xi }) \right\| _{L^q} \\&\le \sum _{\pmb {\beta } \in \{0,1\}^n } \left\| {\mathcal {H}}_{\pmb {\beta }} \left( e^{-i\pmb {\eta }_{\pmb {\beta }}' \pmb {l}_{\pmb {\beta }}}\phi _\Delta ^{\pmb {\alpha }} \big (-\pmb {\eta }_{\pmb {\beta }}-\pmb {\xi }_{\pmb {1}-\pmb {\beta }}\big )g\big (\pmb {\eta }_{\pmb {\beta }}+\pmb {\xi }_{\pmb {1}-\pmb {\beta }} \big ) \right) (\pmb {\xi }) \right\| _{L^q} \\&\le \sum _{\pmb {\beta } \in \{0,1\}^n } C_q^{\pmb {\beta } } \left\| e^{-i\pmb {\eta }_{\pmb {\beta }}' \pmb {l}_{\pmb {\beta }}} \phi _\Delta ^{\pmb {\alpha }} (\pmb {\xi })g (\pmb {\xi }) \right\| _{L^q} \\&\le \sum _{\pmb {\beta } \in \{0,1\}^n } C_q^{\pmb {\beta } } \left\| \phi _\Delta ^{\pmb {\alpha }} (\pmb {\xi })g (\pmb {\xi }) \right\| _{L^q} \\&\le C_q \left\| \phi _\Delta ^{\pmb {\alpha }} (\pmb {\xi })g (\pmb {\xi }) \right\| _{L^q}, \end{aligned}$$

where we set $C_q = 2^n\max _{\pmb {\beta } \in \{0,1\}^n} C_q^{\pmb {\beta } }$.

(2) Set

$$\begin{aligned} q_1 =1, \quad \frac{1}{p_k}+\frac{1}{q_k}=\frac{1}{q_{k-1}},\ k=2,\cdots ,N. \end{aligned}$$

Let

$$\begin{aligned}&C_{norm} = \prod _{i=1}^n2M_ih_i,\ C_3 = C_1\cdot 2^nC_{q_2},\\&C_k = \frac{C_3C_{norm}}{2^{n(k-2)}} \prod _{i=2}^{k}\Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ) \Vert _{L^{p_i}}, k \ge 4. \end{aligned}$$

For $i=1,\cdots , N-1$, we have

$$\begin{aligned}&\left| {\tilde{v}}^{i}_{\pmb {\alpha }}(\pmb {\xi }) - {\hat{v}}^{i}_{\pmb {\alpha }} (\pmb {\xi }) \right| = \frac{e^{-\Delta \psi (i\pmb {\alpha })}}{2^n} \left| {\mathcal {P}}^{\Delta }_{\pmb {h},\pmb {M}}{\tilde{v}}^{i+1}_{\pmb {\alpha }}(\pmb {\xi }) - {\mathcal {P}}^{\Delta }{\hat{v}}^{i+1}_{\pmb {\alpha }} (\pmb {\xi })\right| \\&\quad \le \frac{e^{-\Delta \psi (i\pmb {\alpha })}}{2^n} \left( \left| {\mathcal {P}}^{\Delta }_{\pmb {h},\pmb {M}}{\tilde{v}}^{i+1}_{\pmb {\alpha }}(\pmb {\xi }) - {\mathcal {P}}^{\Delta }{\tilde{v}}^{i+1}_{\pmb {\alpha }} (\pmb {\xi }) \right| + \left| {\mathcal {P}}^{\Delta }\left( {\tilde{v}}^{i+1}_{\pmb {\alpha }} -{\hat{v}}^{i+1}_{\pmb {\alpha }}\right) (\pmb {\xi }) \right| \right) . \end{aligned}$$

Now we estimate $I_1$ and $I_2$.

$$\begin{aligned} I_1&= C_1 \int _{{\mathbb {R}}^n}\left| \phi _{\Delta }^{\pmb {\alpha }}(-\pmb {\xi }) \right| \cdot \left| {\mathcal {P}}^{\Delta }_{\pmb {h},\pmb {M}}{\tilde{v}}^{2}_{\pmb {\alpha }}(\pmb {\xi }) - {\mathcal {P}}^{\Delta }{\tilde{v}}^{2}_{\pmb {\alpha }} (\pmb {\xi }) \right| d\pmb {\xi }\\&\le C_1 \int _{{\mathbb {R}}^n}\left| \phi _{\Delta }^{\pmb {\alpha }}(-\pmb {\xi }) \right| d\pmb {\xi } \cdot \Vert {\mathcal {P}}^{\Delta }_{\pmb {h},\pmb {M}}{\tilde{v}}^{2}_{\pmb {\alpha }} - {\mathcal {P}}^{\Delta }{\tilde{v}}^{2}_{\pmb {\alpha }}\Vert _{L^{\infty }} \\&\le C_2 \Vert {\mathcal {P}}^{\Delta }_{\pmb {h},\pmb {M}}{\tilde{v}}^{2}_{\pmb {\alpha }} - {\mathcal {P}}^{\Delta }{\tilde{v}}^{2}_{\pmb {\alpha }} \Vert _{L^{\infty }}, \end{aligned}$$

where $C_2 = C_1 \Vert \phi _{\Delta }^{\pmb {\alpha }}\Vert _{L^1}$ and

$$\begin{aligned} I_2&= C_1 \int _{{\mathbb {R}}^n}\left| \phi _{\Delta }^{\pmb {\alpha }}(-\pmb {\xi }) \right| \cdot \left| {\mathcal {P}}^{\Delta }\left( {\tilde{v}}^{2}_{\pmb {\alpha }} -{\hat{v}}^{2}_{\pmb {\alpha }}\right) (\pmb {\xi }) \right| d\pmb {\xi } \\&\le C_1 \Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ) \Vert _{L^{p_2}} \cdot \Vert {\mathcal {P}}^{\Delta }\left( {\tilde{v}}^{2}_{\pmb {\alpha }} -{\hat{v}}^{2}_{\pmb {\alpha }}\right) \Vert _{L^{q_2}} \\&\le C_3 \Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ) \Vert _{L^{p_2}} \cdot \Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot )\left( {\tilde{v}}^{2}_{\pmb {\alpha }} -{\hat{v}}^{2}_{\pmb {\alpha }}\right) (\cdot ) \Vert _{L^{q_2}}, \end{aligned}$$

where $C_3 = C_1\cdot 2^nC_{q_2}$. Applying the Holder inquality and the Minkowski inequality, we obtain

$$\begin{aligned}&\Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot )\left( {\tilde{v}}^{2}_{\pmb {\alpha }} -{\hat{v}}^{2}_{\pmb {\alpha }}\right) (\cdot ) \Vert _{L^{q_2}} \nonumber \\&\quad \le \Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ) \Vert _{L^{p_3}} \cdot \Vert {\tilde{v}}^{2}_{\pmb {\alpha }} -{\hat{v}}^{2}_{\pmb {\alpha }}\Vert _{L^{q_3}} \nonumber \\&\quad \le \frac{e^{-\Delta \psi (i\pmb {\alpha })}}{2^n} \Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ) \Vert _{L^{p_3}} \cdot \Vert {\mathcal {P}}^{\Delta }_{\pmb {h},\pmb {M}}{\tilde{v}}^{3}_{\pmb {\alpha }} - {\mathcal {P}}^{\Delta }{\hat{v}}^{3}_{\pmb {\alpha }} \Vert _{L^{q_3}} \nonumber \\&\quad \le \frac{e^{-\Delta \psi (i\pmb {\alpha })}}{2^n} \Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ) \Vert _{L^{p_3}} \left( \Vert {\mathcal {P}}^{\Delta }_{\pmb {h},\pmb {M}}{\tilde{v}}^{3}_{\pmb {\alpha }} - {\mathcal {P}}^{\Delta }{\tilde{v}}^{3}_{\pmb {\alpha }} \Vert _{L^{q_3}} + \Vert {\mathcal {P}}^{\Delta }\left( {\tilde{v}}^{3}_{\pmb {\alpha }} -{\hat{v}}^{3}_{\pmb {\alpha }}\right) (\pmb {\xi })) \Vert _{L^{q_3}} \right) \nonumber \\&\quad \le \frac{\Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ) \Vert _{L^{p_3}} }{2^n} \Vert {\mathcal {P}}^{\Delta }_{\pmb {h},\pmb {M}}{\tilde{v}}^{3}_{\pmb {\alpha }} - {\mathcal {P}}^{\Delta }{\tilde{v}}^{3}_{\pmb {\alpha }} \Vert _{L^{q_3}} + C_{q_3}\Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ) \Vert _{L^{p_3}} \Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot )\left( {\tilde{v}}^{3}_{\pmb {\alpha }} -{\hat{v}}^{3}_{\pmb {\alpha }}\right) (\cdot ) \Vert _{L^{q_3}} \nonumber \\&\quad \le \cdots \le \sum _{k=3}^{N}\frac{1}{2^{n(k-2)}} \prod _{i=3}^{k}\Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ) \Vert _{L^{p_i}} \Vert {\mathcal {P}}^{\Delta }_{\pmb {h},\pmb {M}}{\tilde{v}}^{k}_{\pmb {\alpha }} - {\mathcal {P}}^{\Delta }{\tilde{v}}^{k}_{\pmb {\alpha }} \Vert _{L^{q_k}}\nonumber \\&\qquad + \prod _{k=3}^{N} C_{q_k}\Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ) \Vert _{L^{p_k}} \Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot )\left( {\tilde{v}}^{N}_{\pmb {\alpha }} -{\hat{v}}^{N}_{\pmb {\alpha }}\right) (\cdot )\Vert _{L^{q_N}}. \end{aligned}$$

(A.13)

We point out the inequality

$$\begin{aligned} \Vert f\Vert _{L^{p}} \le C_{norm}\Vert f\Vert _{L^{\infty }}, \end{aligned}$$

(A.14)

which holds for $p>1$ and f supported in the bounded hyperrectangle. We have

$$\begin{aligned}&\Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot )\left( {\tilde{v}}^{N}_{\pmb {\alpha }} -{\hat{v}}^{N}_{\pmb {\alpha }}\right) (\cdot )\Vert _{L^{q_N}}\nonumber \\&\quad \le \Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot )\left( {\tilde{v}}^{N}_{\pmb {\alpha }} -{\hat{v}}^{N}_{\pmb {\alpha }}\right) (\cdot )1_{\Omega _{\pmb {h},\pmb {M}}}(\cdot )\Vert _{L^{q_N}}+\Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot )\left( {\tilde{v}}^{N}_{\pmb {\alpha }} -{\hat{v}}^{N}_{\pmb {\alpha }}\right) (\cdot )1_{\Omega ^c_{\pmb {h},\pmb {M}}}(\cdot )\Vert _{L^{q_N}}\nonumber \\&\quad \le C_{norm}\Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot )\left( {\tilde{v}}^{N}_{\pmb {\alpha }} -{\hat{v}}^{N}_{\pmb {\alpha }}\right) (\cdot )1_{\Omega _{\pmb {h},\pmb {M}}}(\cdot )\Vert _{L^{\infty }} + \Vert {\hat{g}}_{\pmb {\alpha }}\Vert _{L^{\infty }}\Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot )1_{\Omega ^c_{\pmb {h},\pmb {M}}}(\cdot )\Vert _{L^{q_N}}\nonumber \\&\quad \le C_{norm}\Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot )\left( {\tilde{v}}^{N}_{\pmb {\alpha }} -{\hat{v}}^{N}_{\pmb {\alpha }}\right) (\cdot )1_{\Omega _{\pmb {h},\pmb {M}}}(\cdot )\Vert _{L^{\infty }} + \Vert {\hat{g}}_{\pmb {\alpha }}\Vert _{L^{\infty }}I_{\pmb {h},\pmb {M}}. \end{aligned}$$

(A.15)

Let

$$\begin{aligned} C'_N = C_3\max \{C_{norm},\Vert {\hat{g}}_{\pmb {\alpha }}\Vert _{L^{\infty }}\}\Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ) \Vert _{L^{p_2}} \prod _{k=3}^{N} C_{q_k}\Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ) \Vert _{L^{p_k}}. \end{aligned}$$

Using (A.14) and (A.15), we can bound the RHS of (A.13) by the RHS of (3.43).

Next we provide estimates for the constants. First, we have

$$\begin{aligned} \Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ) \Vert _{L^{p}}&= \left( \int _{{\mathbb {R}}^n} |\phi _{\Delta }^{\pmb {\alpha }}(-\pmb {\xi }) |^p d\pmb {\xi } \right) ^{\frac{1}{p}} \le \left( \int _{{\mathbb {R}}^n} a^pe^{-p\Delta b|\pmb {\xi }|^{\nu }} d\pmb {\xi } \right) ^{\frac{1}{p}}\\&\le (p\Delta b)^{-\frac{n}{\nu p}}\left( \int _{{\mathbb {R}}^n}a^pe^{-|\pmb {\eta }|^{\nu }}d\pmb {\eta } \right) ^{\frac{1}{p}},\\&= A^{\frac{1}{p}} p^{-\frac{n}{\nu p}}, \end{aligned}$$

where

$$\begin{aligned} \pmb {\eta } = (p\Delta b)^{\frac{1}{\nu }}\pmb {\xi },\ A = \frac{\int _{{\mathbb {R}}^n}a^pe^{-|\pmb {\eta }|^{\nu }}d\pmb {\eta } }{(\Delta b)^{\frac{n}{\nu }}}. \end{aligned}$$

The constant $C_p$ is given by

$$\begin{aligned}C_p^{\frac{1}{n}} = {\left\{ \begin{array}{ll} \tan (\pi /2p), &{} 1<p\le 2, \\ \cot (\pi /2p), &{} 2 \le p <\infty . \end{array}\right. } \end{aligned}$$

Then, with $q_k = 2-\frac{1}{2^k}$, we obtain (assume $A > 1$, otherwise replace $A^{\frac{1}{p}}$ with 1)

$$\begin{aligned}&p_k = \frac{q_{k-1}q_k}{q_k-q_{k-1}} = 2^k\cdot \left( 2-\frac{1}{2^k} \right) \cdot \left( 2-\frac{1}{2^{k-1}} \right) \ge 2^k , \quad p_k ^{-\frac{1}{ p_k}} \le 2^{-\frac{k}{2^{k}}},\\&\tan \left( \frac{\pi }{2q_k}\right) \approx 1+2\left( \frac{\pi }{2q_k}-\frac{\pi }{4}\right) = 1+\frac{\pi }{2^{k+1}\left( 2-\frac{1}{2^k} \right) } \le 1+\frac{\pi }{2^{k+1}}, \end{aligned}$$

and

$$\begin{aligned}&\prod _{i=2}^{k} \Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ) \Vert _{L^{p_i}} \le \prod _{i=2}^{k} \left( 2^{-\frac{i}{2^{i}}} \right) ^{\frac{n}{\nu }} A^{\frac{1}{2^i}} \le \sqrt{A }, \quad 3\le k \le N, \\&\prod _{k=3}^{N} C_{q_k}\Vert \phi _{\Delta }^{\pmb {\alpha }}(-\cdot ) \Vert _{L^{p_k}} \le \prod _{k=3}^{N} \left( 2^{-\frac{k}{2^{k}}} \right) ^{\frac{n}{\nu }} \cdot A^{\frac{1}{2^k}} \cdot \left( 1+\frac{\pi }{2^{k+1}} \right) ^n \le A^{\frac{1}{4}}e^n. \end{aligned}$$

Thus $C_k \sim O\left( \frac{1}{2^{n(k-2)}} \right)$ and $C'_N \sim O(1)$. $\square$